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As I can understand about the question, it is required to prove,

$$\operatorname{Corr}( X, Y ) + \operatorname{Corr}( X, Z ) + \operatorname{Corr}( Y, Z ) \geq -3/2 \tag i$$

But for any two variables $X,Y$;

$$-1\leq \operatorname{Corr}( X, Y )\leq 1$$

Then how can I obtain the above result (i).?

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    $\begingroup$ Possibly related to stats.stackexchange.com/q/254282/3277 $\endgroup$ – ttnphns Nov 26 '18 at 18:55
  • $\begingroup$ In the link, the sum is solved by setting the value of the determinent 》 0 as R is pd and the one unknown value is determined..But here no value of the elements in the correlation matrix is known..Then how can we find the inequality using that relation? Can you please explain? Thank you. $\endgroup$ – user163580 Nov 26 '18 at 19:11
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    $\begingroup$ Hints: (1) By standardizing the variables, you may assume the correlations are covariances. (2) Compute the variance of $X+Y+Z$ and remember it cannot be negative. $\endgroup$ – whuber Nov 26 '18 at 19:25
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    $\begingroup$ The question is essentially answered in stats.stackexchange.com/q/72790/6633 where the answers show that the average of the three correlations is at least $-\frac 12$ from which it is easy to the result sought here. I vote to close this as a duplicate. $\endgroup$ – Dilip Sarwate Nov 26 '18 at 21:04
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\begin{align} & \operatorname{Var}(A+B+C) \\[8pt] = {} & \operatorname{Var}(A)+\operatorname{Var}(B) + \operatorname{Var}(C) + 2\operatorname{Cov}(A,B)+2\operatorname{Cov}(B,C) + 2\operatorname{Cov}(A,C) \\[8pt] \ge {} & 0 \end{align}

Now let $A=X/\operatorname{std}(X)$ and similar for $B$ and $C$, so we get

$3+2\operatorname{Corr}(X,Y)+2\operatorname{Corr}(Y,Z)+2\operatorname{Corr}(X,Z) \ge 0$ as desired.

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